Chapter Eight. We've had a good look at power and power factor, we're now going to look into power factor correction and the reason for it and how to correct it. Now, in order to get a visual on power factor and power factor correction, I'm gonna hypothesize a utility setup here for the moment I'm in and I'm going to look at a generating station. And just for the sake of argument, I'm going to assume that that generating station is a hydraulic generating station. In other words, the prime mover for spinning the generator is water movement through the generator. And I'm going to imagine that initially when the utility First puts the generator into operation, that it has a fairly light load.

And that light load is basically a resistive load. And it's symbolized by the light bulbs there. And and that's a little bit hypothetical I know. But let's assume that we have a fairly consistent resistive load on the generator. So I've indicated the current flowing out of the generator with i G or I subscript G and then the generator feeds a transformer that has current flowing through the transformer of I subscript T, and the transformer then feeds that current and power into a line that ultimately feeds the resistive load. And that line is carrying a current of I subscript L. So if we were to take them measurement of the average power flowing to that resistive load, it's made up of the voltage times the amperage, and I've indicated a phasor for the power average in the upper right hand corner there, and it's the burgundy arrow with basically zero power factor angle.

So just multiplying the current and the voltage together will give us the average power as long as you remain taking those measurements and RMS values. Now, as evolution begins on this utility power station, I'm going to add another load and the load that I'm adding is primarily a motor load, which is an inductive motor load. And let's for argument's sake, just say that it's it is an unloaded set of motors. In other words, they're idle. And I'm doing that just to make a demonstration here. So the amount of power flowing to the generator or flowing from the generator, I should say to both the motor and the light bulbs hasn't changed the, the average power is the same, there's a certain amount of reactive power that is being supplied to energize the motors as they're running on no load.

However, the power the average power consumption is the same, and hence the amount of water that's flowing through the generator doesn't change. So the power that's being delivered to both loads is the same as if the motor load wasn't there except for the reactive part of it. Now the reactive part of an inductive load if you remember the power the power flowing to the inductive load is half of the cycle is supplying power to the load and the other half of the cycle, the motor is supplying power back into the system. However, there is an increase in current, because we have to supply the reactive power. And you can see that the VA or the apparent power is larger because we are supplying a certain amount of reactive load. Now, let's for argument's sake add another load, pretty much the same as we just added.

And it is for argument's sake, let's assume that it's a motor load as well. And that's going to increase the amount of apparent power that's flowing. And again, let's assume that the motor is unloaded. So it's basically adding only reactive a reactive element to the load but You can see from our phasor diagram that the apparent power has increased. Now the voltage coming from the generator hasn't changed, it's still the same, whatever voltage level it is. So because the apparent power is made up of the voltage times the current, then the only thing that's increased is the current.

So you can see there's a fairly hefty increase in the current from the generator, the current through the transformer. And the current through the line to supply are three loads that we have there. Now let's add one more load to this scenario. And again, let's assume that it's a motor load that's unloaded. And the reactive element certainly goes up because we're supplying the reactive element or the The the magnetic field on the on the rotors of the transformer is increasing or the stators of the motor. Anyway, the the inductive load is increasing however we've said all the motors are unloaded, so we're not increasing the average real power that's supplying a load only the reactive power Hence, the apparent power is increasing sequentially as we add these motor loads, however, the average power is that the real power is still the same.

But you can see that the current being supplied by the generator and the transformer in the line has increased substantially. So most of the time, the utilities are billing on real power consumption, the amount of energy that's flowing to this system and the amount of Real energy and power that's flowing to the system is related to the amount of water that's flowing through the generator, which hasn't changed, it's still the same amount of water that's flowing because we're only supplying the lighting load as far as the real load is consulted concerned. So the utilities, you really don't like to supply all of this current because you can imagine that the transformer and the generator and the lines have to have the capacity to supply this load. And if they're only going to get revenue from the real power that's being supplied, it's basically unfair. So, utilities will penalize the large inductive loads, which usually are industrial loads, if they happen to have a large component of inductance or reactive power use because it has supply that reactive power anyway in the form of in the form of current that's being flown to the load.

So the the utility has a way of trying to discourage this. And that is a they can build the customer for, but the customer has an opportunity to counteract that inductive load that they're supplying. And they do that because all of this load the reactive load that the motors are bringing into the system is inductive. So if we supply an equal amount of capacitive type load, that would counteract the inductive part of the load, and that's exactly what happens. For example, they will on the feeders supplying the motor load, they can apply a certain amount of capacitive capacitors to the load which will counteract the reactive load going to the inductive element. So now the generator will supply the energy to the motor as well as the lighting, but the reactive element will transfer every half cycle between the capacitor and the motor loads.

So the reactive element is being satisfied by an exchange between the capacitors and the motor load. So are the amount of parent power that's being supplied by the generator drops down and hence the current from the generator, the transformer and the line drops down. So we get a better power factor and we say better that means it's a smaller power factor angle. So let's take this another step. Let's say we add some more capacitance this time To the next load, and what happens is that capacitor will counteract the reactive element of our power triangle and you will get the capacitor trading off reactive power between the motor and the capacitor as time goes on, and we haven't added any more generation to the system, we've just added capacitance and that capacitance will drop the demand for current or the apparent power that has to be supplied by the generator.

Now we got one more motor we could add another set of capacitor banks to that load and that load would drop the apparent power down one more step. And now all three of the motor loads are being supplied as far as their reactive power is concerned by the company fasteners that capacitors are trading the reactive power back and forth to the motors. And the generator just sees the whole load system as unity power factor if the capacitors are sized properly to counteract the inductive load, and there's a way of calculating that, and we're going to have a look at that in the next few slides. We've just had a look at a hypothetical case of adding capacitors to relatively high inductive load to counteract the inductance of the motors themselves. And adding that capacitance will improve the power factor. And this is what we call power factor correction.

And we're going to demonstrate that now by having a look at an example. Let's assume that we have a, an inductive resistive load, it could be a motor load, but we're representing it by an inductor in series with a resistor. And we have a source voltage, a sinusoidal source voltage of E s. And this circuit has a source voltage of 220 volts AC. And it's a frequency is 60 cycles which it is in North America. And this power factor happens to be 0.82. And it's drawing a real power of 5.4 kilowatts.

What we would like to do is correct the power factor or improve the power factor by adding a capacitance or a passive capacitive load, but we want to know how much capacitance must be added to improve The power factor to 0.96 we'll just move those quantities down to the bottom right hand corner so we can keep track of them during our calculations. And you can see the power triangle is made up of the real power plus the reactive power to give us the apparent power, which is p plus Q is equal to s, and the power factor angle is theta, which is the contained angle between the P and the s. Now, we're going to have a come up with a new power factor and a new apparent power and new power factor angle. So I'm going to relabel the S q and theta of our original power factor, triangle as s one q one and theta one, the power, the real power does not change it's going to be the same For both the new power factor and the old one, and the new power factor, or power triangle will be what you see in blue there, the apparent power is going to be s to the new contained angle is theta two.

And the new reactive power is given by q two. We know what the original power factor is 0.82, which is given by taking the cosine of the original contained angle theta one. So, if we take the inverse of the power factor that would give us our angle theta one. So, the inverse of 0.82 works out to almost 35 degrees 34.9152. Now, the power factor itself is given by taking p all over s one or the real Power all over the apparent power of our original power triangle, but we don't know what s one is, but we do know what the power factor is. So if we want to find out what s one is, we just do a mathematical shift of that equation and we can find s one by dividing the real power by the power factor, and that is 5400 or 5.4 kilowatts divided by point eight, two, which gives us 6585.

So, if we want to find out what q one is, all we got to do is take the sign of the a the theta one and multiply it by the apparent power s one and that will give us our value for q1. And that works out to 3769. Remember now that the objective was to improve the power factor of this circuit here by adding capacitance, and as such we'll come up with a new power triangle and that power triangle will be this one in blue here, and we have to find out what the values of s two q2, and theta two are. theta two we can easily calculate because we were the objective was to come up with a new power factor, which is point nine six. So if we just take the inverse of that power factor, we'll come up with a new power factor angle which is 16.2602 degrees.

And in calculating the new apparent power, it's using the same logic that we use for calculating the apparent power of the original power triangle. And that was to take the real power and divide it by the new power factor. And that it's that is 5400, all over point nine six, which works out to 5625. We now can calculate what q2 is or the reactive component of the power triangle by taking the sign of the new power factor angle and multiplying it by the new apparent power, or s two times sine theta two, which equals 5625 times the sine of 16.2602 degrees which is 1000. 575 now in adding the capacitance to the circuit, we are going to add a certain amount of reactive component. And the amount of reactive component that we're going to add is going to be the amount that is required to reduce the original q1 down to q2.

And we're going to call that q c which is the reactance of the capacitive or capacitance. And we can now introduce this equation, which is the original reactance q1 minus or sorry, plus the reactance of the capacitor q c, which is gonna turn out to be negative because it's going in the other direction, but we'll just confirm that it is a vector right now and we're going to take vector q1 and add vector To see, and we're going to end up with vector q2. We know what q1 and q2 are. So we can just take the q1 to the other side of the equation and we can solve for q C, and q2 minus q1 is just 1575 minus 3769, which works out to minus two to minus 2194. Now, we also remember that the Q c is also equal to the voltage the square of the voltage across it the magnitude of the square the voltage across it all over j times x conjugate where x is the reactance of the capacitor.

And if we substitute in the reactance of the capacitor, we now have the x magnitude squared all over j one all over two pi f c, which it equates to minus j two pi fc times the EC EAX magnitude squared. So this is the equation for the phasor QC. But we're only interested in its magnitude because we can calculate the capacitance from the magnitude of QC. And we know that the magnitude of QC is going to be two pi fc times the voltage drop across the capacitor squared. Now we know the voltage drop across the capacitor is equal to the source voltage. So we can We know what the source voltage is, so we know what EAX is.

So we can rewrite this equation because we know everything except the capacitance in that equation. So let's take the cut the c value or the capacitance to the left hand side of the equation and bring it all the other values to the right hand side of the equation. And we will come up with this where the capacitance is equal to the magnitude of QC all over two pi f times the square of the source voltage. And substituting the numbers for that we know that the magnitude of QC is 2194. And we know what two pi is, of course, and we know what the frequency is 60 cycles and we know what the source voltage is is 220 volts. And we can square that.

And if you do the calculations that comes out to 120 microfarads which is what we were looking to calculate in the first place. This ends the chapter